The motivation for this question comes from a proof I just saw demonstrating that a circle in the plane and $[a,b]$ are not homeomorphic. The argument consists of removing a point from the circle, and also a point from $[a,b]$ which i not an endpoint. Upon removal of this point from both we observe that the circle remains connected while the interval is disconnected.
From my understanding, $\mathbb{R}$ is connected and therefore has no proper clopen subsets. But if we can take $[a,b] \in \mathbb{R}$ and disconnect it by removing a point which is neither $a$ or $b$, then this subset of $\mathbb{R}$ has a proper clopen subset, which means that $\mathbb{R}$ has a proper clopen subset. I know that this reasoning is somehow wrong, but I'm not sure where my misunderstanding lies and I would appreciate any help.
Also, what would be the proper clopen subset of $[a,b]$ minus a single point that makes it disconnected?
Best Answer
Note that a set $U \subset [a, b]$ is open in $[a, b]$ if $U = [a, b] \cap V$ for some open set $V$ in $\mathbb R$. The term "open in $[a, b]$" is not the same as "open in $\mathbb R$". Indeed, for each $c\in (a, b)$, $[a, c)$ is open in $[a, b]$ but not open in $\mathbb R$. Thus a clopen set in $[a, b]$ might not be a clopen set in $\mathbb R$